Solve {l}{x^3+27=(x+3)(x^2-}{(a+1)(a^2-a+1)=}

$$\left. \begin{array} { l } { x ^ { 3 } + 27 = ( x + 3 ) ( x ^ { 2 } - } \\ { ( a + 1 ) ( a ^ { 2 } - a + 1 ) = } \end{array} \right.$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{23}i+3}{2}\approx 1.5-2.397915762i$$

$$x=-3$$

$$x=\frac{3+\sqrt{23}i}{2}\approx 1.5+2.397915762i$$

Solution Steps

Subtract x from both sides.

$$x^{3}+27-x=3$$

Subtract 3 from both sides.

$$x^{3}+27-x-3=0$$

Subtract 3 from 27 to get 24.

$$x^{3}+24-x=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{3}-x+24=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 24 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±24,±12,±8,±6,±4,±3,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-x+24 by x+3 to get x^{2}-3x+8. Solve the equation where the result equals to 0.

$$x^{2}-3x+8=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -3 for b, and 8 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 8}}{2}$$

Do the calculations.

$$x=\frac{3±\sqrt{-23}}{2}$$

Solve the equation x^{2}-3x+8=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{23}i+3}{2} x=\frac{3+\sqrt{23}i}{2}$$

List all found solutions.

$$x=-3 x=\frac{-\sqrt{23}i+3}{2} x=\frac{3+\sqrt{23}i}{2}$$

Solve for x

$$x=-3$$

Solution Steps

Subtract x from both sides.

$$x^{3}+27-x=3$$

Subtract 3 from both sides.

$$x^{3}+27-x-3=0$$

Subtract 3 from 27 to get 24.

$$x^{3}+24-x=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{3}-x+24=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 24 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±24,±12,±8,±6,±4,±3,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-x+24 by x+3 to get x^{2}-3x+8. Solve the equation where the result equals to 0.

$$x^{2}-3x+8=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -3 for b, and 8 for c in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 8}}{2}$$

Do the calculations.

$$x=\frac{3±\sqrt{-23}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=-3$$